C3 chapter 6 - trigonometry

In this chapter you will learn

about the reciprocal and inverse trigonometric ratios and how to solve equations and prove identities involving them.
This topic requires a sound recall of basic trigonometry, Pythagoras' theorem, an understanding of indices including negative indices and the ideas of domain and range and inverse functions.

Here's a question or two to get us started:
We know that:
${\displaystyle} x^{3} = x {\times} x {\times} x$
${\displaystyle} x^{2} = x {\times} x$
and
${\displaystyle} x^{1} = x$
and so, following this logic we arrive at:
${\displaystyle} x^{-1} = \dfrac{1}{x}$


Similarly if
${\displaystyle} \sin^{3}{\theta} = \sin{\theta} {\times} \sin{\theta} {\times} \sin{\theta}$
${\displaystyle} \sin^{2}{\theta} = \sin{\theta} {\times} \sin{\theta}$
and
${\displaystyle} \sin^{1}{\theta} = \sin{\theta}$
...then you'd think:
${\displaystyle} \sin^{-1}{\theta} = \dfrac{1}{\sin{\theta}}$
*So is it?

But we also know that when we look at the function
${\displaystyle} f(x): x \to x^{3}$
we get the inverse function such that:
${\displaystyle} f^{-1}(x): f(x) \to x$
...which is to say
${\displaystyle} f^{-1}(x): x \to \sqrt[3]{x}$
And the function:
${\displaystyle} g(x): x \to -x$
has its inverse function which gets from the output back to the input
${\displaystyle} g^{-1}(x): g(x) \to x$
and on this occasion that inverse function is actually the same as the original (why?)
${\displaystyle} g^{-1}(x): x \to -x$
So you'd hope that the function:
${\displaystyle} \sin^{-1}{\theta}$
would be the inverse of the function
${\displaystyle} \sin{\theta}$
*...so is it that then?

And then there's the killer question: are these both correct, and if not, which if either is?

Sections

C3 section 6.1

Recall how sine and cosine are related to the unit circle:




Now notice that the other lengths are also useful:




How can we be sure that the lengths marked are what they claim to be?
Tangere is Latin for the present active infinitive of tangō. "to touch, to grasp" (1). Why is the tangent called that?
Secāre is the Latin present active infinitive of secō. (2) It means to cut or divide. Why is the secant called that?

http://en.wiktionary.org/wiki/tangerehttp://en.wiktionary.org/wiki/seco#Latin

You need to be able to:

You need to know:

Examples:

C3 Exercise 6A

• C3 Exercise 6A worked solutions:
• Questions similar to this exercise:

C3 section 6.2

C3 Exercise 6B

• C3 Exercise 6B worked solutions:
• Questions similar to this exercise:

C3 section 6.3

Prior knowledge: you need a secure recall of the meanings of sec(θ), cosec(θ) and cot(θ). You should also be fluent with the definition of tan(θ) in terms of sin(θ) and cos(θ). You must be able to solve simple trigonometric equations, including those with double angles and solve quadratics both by factorisation and completing the square.

L.O. WALT:

• simplify expressions involving sec(θ), cosec(θ) and cot(θ)
• prove identities involving sec(θ), cosec(θ) and cot(θ)
• solve equations involving sec(θ), cosec(θ) and cot(θ).

Examples:

C3 Exercise 6C

• C3 Exercise 6C worked solutions (mostly complete):
  

  C3 Exercise 6C (reciprocal trig. identities & equations).jnt
  

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  C3 Exercise 6C (reciprocal trig etc).pdf
  

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• Questions similar to this exercise:

C3 section 6.4

Prior knowledge: you need a secure recall of the work above on sec(θ), cosec(θ) and cot(θ). You should also be familiar with the identity sin²θ + cos²θ = 1. You must be able to use the 'CAST diagram' to determine the sign of trig ratios and use Pythagoras' theorem and right-triangles to derive unknown trig ratios from known facts. You will need to be able to spot instances of the difference of two squares including those disguised in higher powers and other functions.

L.O. WALT:

• prove the identities involving sec²(θ), cosec²(θ), tan²(θ) and cot²(θ)
• find exact values using these identities and other methods
• prove a wider variety of reciprocal trig identities
• solve equations involving powers of sec(θ), cosec(θ) and cot(θ).

Examples:

C3 Exercise 6D

• C3 Exercise 6D worked solutions (partially complete):

C3 Exercise 6D (trig proof).jnt

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C3 Exercise 6D (trig proof) [partly done].pdf

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C3 section 6.5